3.11.17 \(\int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx\) [1017]

Optimal. Leaf size=63 \[ \frac {b B x}{e^2}-\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {(2 b B d-A b e-a B e) \log (d+e x)}{e^3} \]

[Out]

b*B*x/e^2-(-a*e+b*d)*(-A*e+B*d)/e^3/(e*x+d)-(-A*b*e-B*a*e+2*B*b*d)*ln(e*x+d)/e^3

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \begin {gather*} -\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {\log (d+e x) (-a B e-A b e+2 b B d)}{e^3}+\frac {b B x}{e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(A + B*x))/(d + e*x)^2,x]

[Out]

(b*B*x)/e^2 - ((b*d - a*e)*(B*d - A*e))/(e^3*(d + e*x)) - ((2*b*B*d - A*b*e - a*B*e)*Log[d + e*x])/e^3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(a+b x) (A+B x)}{(d+e x)^2} \, dx &=\int \left (\frac {b B}{e^2}+\frac {(-b d+a e) (-B d+A e)}{e^2 (d+e x)^2}+\frac {-2 b B d+A b e+a B e}{e^2 (d+e x)}\right ) \, dx\\ &=\frac {b B x}{e^2}-\frac {(b d-a e) (B d-A e)}{e^3 (d+e x)}-\frac {(2 b B d-A b e-a B e) \log (d+e x)}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 56, normalized size = 0.89 \begin {gather*} \frac {b B e x-\frac {(b d-a e) (B d-A e)}{d+e x}+(-2 b B d+A b e+a B e) \log (d+e x)}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(A + B*x))/(d + e*x)^2,x]

[Out]

(b*B*e*x - ((b*d - a*e)*(B*d - A*e))/(d + e*x) + (-2*b*B*d + A*b*e + a*B*e)*Log[d + e*x])/e^3

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Maple [A]
time = 0.07, size = 70, normalized size = 1.11

method	result	size
default	\(\frac {b B x}{e^{2}}-\frac {A a \,e^{2}-A b d e -B a d e +B b \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {\left (A b e +B a e -2 B b d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(70\)
norman	\(\frac {\frac {B b \,x^{2}}{e}-\frac {A a \,e^{2}-A b d e -B a d e +2 B b \,d^{2}}{e^{3}}}{e x +d}+\frac {\left (A b e +B a e -2 B b d \right ) \ln \left (e x +d \right )}{e^{3}}\)	\(75\)
risch	\(\frac {b B x}{e^{2}}-\frac {A a}{e \left (e x +d \right )}+\frac {A b d}{e^{2} \left (e x +d \right )}+\frac {B a d}{e^{2} \left (e x +d \right )}-\frac {B b \,d^{2}}{e^{3} \left (e x +d \right )}+\frac {\ln \left (e x +d \right ) A b}{e^{2}}+\frac {\ln \left (e x +d \right ) B a}{e^{2}}-\frac {2 \ln \left (e x +d \right ) B b d}{e^{3}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(B*x+A)/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

b*B*x/e^2-(A*a*e^2-A*b*d*e-B*a*d*e+B*b*d^2)/e^3/(e*x+d)+1/e^3*(A*b*e+B*a*e-2*B*b*d)*ln(e*x+d)

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Maxima [A]
time = 0.29, size = 75, normalized size = 1.19 \begin {gather*} B b x e^{\left (-2\right )} - {\left (2 \, B b d - B a e - A b e\right )} e^{\left (-3\right )} \log \left (x e + d\right ) - \frac {B b d^{2} + A a e^{2} - {\left (B a e + A b e\right )} d}{x e^{4} + d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^2,x, algorithm="maxima")

[Out]

B*b*x*e^(-2) - (2*B*b*d - B*a*e - A*b*e)*e^(-3)*log(x*e + d) - (B*b*d^2 + A*a*e^2 - (B*a*e + A*b*e)*d)/(x*e^4
+ d*e^3)

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Fricas [A]
time = 0.47, size = 102, normalized size = 1.62 \begin {gather*} -\frac {B b d^{2} - {\left (B b x^{2} - A a\right )} e^{2} - {\left (B b d x + {\left (B a + A b\right )} d\right )} e + {\left (2 \, B b d^{2} - {\left (B a + A b\right )} x e^{2} + {\left (2 \, B b d x - {\left (B a + A b\right )} d\right )} e\right )} \log \left (x e + d\right )}{x e^{4} + d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^2,x, algorithm="fricas")

[Out]

-(B*b*d^2 - (B*b*x^2 - A*a)*e^2 - (B*b*d*x + (B*a + A*b)*d)*e + (2*B*b*d^2 - (B*a + A*b)*x*e^2 + (2*B*b*d*x -
(B*a + A*b)*d)*e)*log(x*e + d))/(x*e^4 + d*e^3)

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Sympy [A]
time = 0.26, size = 71, normalized size = 1.13 \begin {gather*} \frac {B b x}{e^{2}} + \frac {- A a e^{2} + A b d e + B a d e - B b d^{2}}{d e^{3} + e^{4} x} + \frac {\left (A b e + B a e - 2 B b d\right ) \log {\left (d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)**2,x)

[Out]

B*b*x/e**2 + (-A*a*e**2 + A*b*d*e + B*a*d*e - B*b*d**2)/(d*e**3 + e**4*x) + (A*b*e + B*a*e - 2*B*b*d)*log(d +
e*x)/e**3

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Giac [A]
time = 2.08, size = 116, normalized size = 1.84 \begin {gather*} {\left (x e + d\right )} B b e^{\left (-3\right )} + {\left (2 \, B b d - B a e - A b e\right )} e^{\left (-3\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - {\left (\frac {B b d^{2} e}{x e + d} - \frac {B a d e^{2}}{x e + d} - \frac {A b d e^{2}}{x e + d} + \frac {A a e^{3}}{x e + d}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(B*x+A)/(e*x+d)^2,x, algorithm="giac")

[Out]

(x*e + d)*B*b*e^(-3) + (2*B*b*d - B*a*e - A*b*e)*e^(-3)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - (B*b*d^2*e/(x*e
+ d) - B*a*d*e^2/(x*e + d) - A*b*d*e^2/(x*e + d) + A*a*e^3/(x*e + d))*e^(-4)

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Mupad [B]
time = 1.09, size = 75, normalized size = 1.19 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{e^3}-\frac {A\,a\,e^2+B\,b\,d^2-A\,b\,d\,e-B\,a\,d\,e}{e\,\left (x\,e^3+d\,e^2\right )}+\frac {B\,b\,x}{e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x))/(d + e*x)^2,x)

[Out]

(log(d + e*x)*(A*b*e + B*a*e - 2*B*b*d))/e^3 - (A*a*e^2 + B*b*d^2 - A*b*d*e - B*a*d*e)/(e*(d*e^2 + e^3*x)) + (
B*b*x)/e^2

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